The original problem is:If a, b, c, d are the position vectors of points A, B, C, D respectively such that ( a → − d → ) . ( b → − c → ) = ( b → − d → ) . ( c → − a → ) = 0then prove that D is the orthocentre of Δ ABC.How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.

Question

The original problem is:If a, b, c, d are the position vectors of points A, B, C, D respectively such that  (            a      →        −            d      →        )  .  (            b      →        −            c      →        )  =  (            b      →        −            d      →        )  .  (            c      →        −            a      →        )  =  0then prove that D is the orthocentre of       Δ   ABC.How do we go about proving that a point is the orthocentre of a triangle? I&#039;ve tried expanding the dot product but I don&#039;t seem to get anywhere.

kuglatid4 · Accepted Answer

That formula states that   A  D  ⊥  B  C (so that   D is on the altitude from   A to   B  C) and that   A  C  ⊥  B  D (so that   D is on the altitude from   B to   A  C). As   D is on two altitudes of the triangle, it is its orthocentre.